s domain analysis phân tích mạch trong miền s

S-Domain Analysis


s-Domain Circuit Analysis
Time domain
(t domain)
Linear
Circuit
Differential
equation

Complex frequency
domain (s domain)
Laplace Transform
L
Laplace Transform
L

Classical
techniques
Response
waveform

Transformed
Circuit
Algebraic
equation
Algebraic
techniques

Inverse Transform

L-1

Response
transform


Kirchhoff’s Laws in s-Domain
t domain

s domain
i2 (t )

Kirchhoff’s current law (KCL)

i1 (t )

i3 (t )

i4 (t )

i1 (t ) + i2 (t ) − i3 (t ) + i4 (t ) = 0

I1 ( s ) + I 2 ( s ) − I 3 ( s ) + I 4 ( s ) = 0
+ v2 (t ) −

Kirchhoff’s voltage law (KVL)

− v1 (t ) + v2 (t ) + v3 (t ) = 0

+ v4 (t ) −

+

+

+

v1 (t )

v3 (t )

v5 (t )

−

−

−

− V1 ( s ) + V2 ( s ) + V3 ( s ) = 0


Signal Sources in s Domain
t domain

s domain
i (t )
+

v(t )

v(t ) = vS (t )
i (t ) = depends _
on circuit

vS (t )

L

V (s )

V ( s ) = VS ( s )
I ( s ) = depends
on circuit

I (s )

_
+

_

i (t ) = iS (t )
v(t )
v(t ) = depends +
on circuit

VS (s )

_

i (t )

Current Source:

Voltage Source:

+
_
+

Voltage Source:

I (s )

_

iS (t )

L

V (s )
+

Current Source:
I S (s)

I ( s ) = VS ( s )
V ( s ) = depends
on circuit

Time and s-Domain Element Models
Impedance and Voltage Source for Initial Conditions

Resistor:
vR (t ) = RiR (t )

Inductor:
diL (t )
vL (t ) = L
dt

Capacitor:

iR (t )

s-Domain

I R (s )

+

+

vR (t )

R

L

_

VR (s )

R

_

iL (t )

I L (s )

+

vL (t )
_

L

L

iC (t )

+
1 t
vC (t ) = ∫ iC (τ )dτ
vC (t )
C 0
_
+ vC (0)

Ls

VL (s )
_

_
+

I C (s )

L

VC (s )
_

VR ( s ) = RI R ( s )

LiL (0)

VL ( s ) = LsI L ( s ) −
LiL (0)

Capacitor:

+

C

Resistor:

Inductor:

+

1

_
+

Time Domain

1 Cs VC ( s ) =
I C ( s) +
Cs
vC (0)
vC( 0 )
s

s


Impedance and Voltage Source for Initial
Conditions
• Impedance Z(s)
Z ( s) =

voltage transform
current transform

with all initial conditions set to zero
• Impedance of the three passive elements
Z R ( s) =

VR ( s )
=R
I R (s)
VL ( s )
= Ls
I L (s)

with iL (0) = 0

VC ( s ) 1
=
I C ( s ) Cs

with vC ( 0 ) = 0

Z L ( s) =
Z C (s) =


Time and s-Domain Element Models
Admittance and Current Source for Initial Conditions
Time Domain
Resistor:

1
iR (t ) = vR (t )

R

Inductor:

iR (t )
+

+

vR (t )

iC (t ) = C

dvC (t )
dt

R

L

_

VR (s )

Resistor:

R

iL (t )

I L (s )

+

L

VL (s )
Ls
_

iC (t )

I C (s )

+

vC (t )
_

Inductor:

+

L

+

C

L

VC (s )
_ 1 Cs

1
VR ( s )
R

I R (s) =

_

1 t
iL (t ) = ∫ vL (τ ) dτ
vL (t )
L 0
_
+ iL (0)

Capacitor:

s-Domain

I R (s )

I L (s) =
i L ( 0)
s

1

VL ( s ) +
Ls
i L( 0 )
s

Capacitor:
I C ( s ) = CsVC ( s ) −
CvC (0)

CvC (0)


Admittance and Current Source for Initial
Conditions
• Admittance Y(s)
Y ( s) =

current transform
1
=
voltage transform Z ( s)

with all initial conditions set to zero
• Admittance of the three passive elements
YR ( s ) =

I R ( s) 1
=
VR ( s ) R

YL ( s ) =

I L ( s) 1
=
VL ( s ) Ls

with iL (0) = 0

YC ( s ) =

I C ( s)
= Cs
VC ( s )

with vC( 0 ) = 0


Example: Solve for Current Waveform i(t)
R

L

i (t )

L

VA
s

+ VR (s ) −

_
+

_
+

V Au (t )

R

I (s )

Ls
_
+

+

VL (s )
LiL (0) _

VA
By KVL: −
+ VR ( s ) + VL ( s ) = 0
s
Resistor: VR ( s ) = RI ( s )
Inductor: VL ( s ) = LsI ( s ) − LiL (0)
V
− A + RI ( s) + LsI ( s) − LiL (0) = 0
s

VA L
iL (0)
+
I (s) =
s ( s + R L) s + R L
VA R VA R
i L ( 0)
=
−
+
s
s+R L s+R L
R
− t
V A V A − RL t
Inverse Transform: i (t ) =  − e + iL (0)e L u (t )

R R
forced response

natural response


Series Equivalence and Voltage Division
I1 ( s )
I (s )

Rest
of
Circuit

+ V1 ( s ) −

Z1
+
V (s)
−

V1 ( s ) = Z1 ( s ) I1 ( s ) = Z1 ( s ) I ( s )

+
Z2 V2 ( s) I 2 ( s )
−

V2 ( s ) = Z 2 ( s ) I 2 ( s ) = Z 2 ( s ) I ( s )

KVL: V ( s ) = V1 ( s ) + V2 ( s )

= ( Z1 ( s ) + Z 2 ( s )) I ( s )
I (s )

Rest
of
Circuit

+
V (s) Z
EQ
−
Z EQ = Z1 + Z 2

Z EQ ( s ) = Z1 ( s ) + Z 2 ( s )

V1 ( s) =

Z1 ( s )
V (s)
Z EQ ( s )

V2 ( s ) =

Z 2 (s)
V (s)
Z EQ ( s)


Parallel Equivalence and Current Division
I1 ( s ) = Y1 ( s )V ( s )

I (s )

Rest
of
Circuit

+
I1 ( s )
I 2 (s)
V (s) Y
Y2

1
−

I 2 ( s ) = Y2 ( s )V ( s )

KCL: I ( s ) = I1 ( s ) + I 2 ( s )

= (Y1 ( s ) + Y2 ( s ))V ( s )
I (s )

Rest
of
Circuit

+
V (s) Y
EQ
−
YEQ = Y1 + Y2

YEQ ( s ) = Y1 ( s ) + Y2 ( s )

I1 ( s ) =

Y1 ( s )
I ( s)
YEQ ( s )

I 2 (s) =

Y2 ( s)
I (s)
YEQ ( s )


Example:
Equivalence Impedance and Admittance
A

L

v1 (t )

R

_
+

B

L
A

V1 ( s )
B

Z EQ ( s ) = Ls + Z EQ1 ( s) = Ls +

Ls

Z EQ ZR
EQ1

Z EQ

Inductor current = 0
at t = 0
+
capacitor voltage = 0
C v2 (t ) Find equivalent impedance at A and B
_
Solve for v2(t)
RCs + 1
1
1
= + Cs =
YEQ1 ( s ) =
Z EQ1 ( s ) R
R

Z EQ1

RLCs 2 + Ls + R
+
=
RCs + 1
1
V2 ( s )
Z EQ1 ( s )
Cs

_
V2 ( s ) =
V1 ( s )
Z EQ
R
=
V1 ( s)
2
RCLs + Ls + R

R
RCs + 1

_
+


General Techniques for s-Domain Circuit
Analysis
• Node Voltage Analysis (in s-domain)
–
–
–
–

Use Kirchhoff’s Current Law (KCL)
Get equations of node voltages
Use current sources for initial conditions
Voltage source
current source

• Mesh Current Analysis (in s-domain)
–
–
–
–

Use Kirchhoff’s Voltage Law (KVL)
Get equations of currents in the mesh
Use voltage sources for initial conditions
Current source
voltage source
(Works only for “Planar” circuits)


Formulating Node-Voltage Equations
Step 0: Transform the circuit into the s domain using current
sources to represent capacitor and inductor initial conditions
Step 1: Select a reference node. Identify a node voltage at each
of the non-reference nodes and a current with every element
in the circuit
Step 2: Write KCL connection constraints in terms of the
element currents at the non-reference nodes
Step 3: Use the element admittances and the fundamental
property of node voltages to express the element currents in
terms of the node voltages
Step 4: Substitute the device constraints from Step 3 into the
KCL connection constraints from Step 2 and arrange the
resulting equations in a standard form

Example: Formulating Node-Voltage Equations
L

iS (t )

R

C

Step 0: Transform the circuit into the s domain using
current sources to represent capacitor and
inductor initial conditions
Step 1: Identify N-1=2 node voltages and a current
with each element

t domain

VA (s )
I 2 ( s)

I S (s)
Reference
node

Step 2: Apply KCL at nodes A and B:
iL (0)
−
− I1 ( s ) − I 2 ( s ) = 0
I

s
Node
A
:
(
)
L
S
s
iL (0)
i (0)
+ I1 ( s ) − I 3 ( s ) = 0
Node B : CvC (0) + L
s
s
Step 3: Express element equations in terms of node
Ls
VB (s )
voltages
1
I1 ( s ) I 3 ( s )
I1 ( s ) = YL ( s )[VA ( s ) − VB ( s )] = [VA ( s ) − VB ( s )]
Ls
1
R
Cs CvC (0) I 2 ( s ) = YR ( s)VA ( s ) = GVA ( s) where G = 1 R
s domain

I 3 ( s ) = YC ( s )VB ( s) = CsVB ( s )

Formulating Node-Voltage Equations (Cont’d)
Step 2: Apply KCL at nodes A and B:
iL (0)
− I1 ( s ) − I 2 ( s ) = 0
Node A : I S ( s ) −
s
i (0)
+ I1 ( s ) − I 3 ( s ) = 0
Node B : CvC (0) + L
s
Step 3: Express element equations in terms of node voltages
1
I1 ( s ) = YL ( s)[VA ( s) − VB ( s)] = [VA ( s ) − VB ( s )]
Ls
I 2 ( s ) = YR ( s )VA ( s) = GVA ( s ) where G = 1 R
I 3 ( s ) = YC ( s )VB ( s ) = CsVB ( s)

Step 4: Substitute eqns. in Step 3 into eqns. in Step 2 and collect
common terms to yield node-voltage eqns.
i (0)
1 
 1 

Node A :  G + VA ( s ) −  VB ( s ) = I S ( s ) − L
s
Ls 
 Ls 

i (0)


 1
 1 
Node B : −  VA ( s ) +  + Cs VB ( s) = CvC (0) + L
s

 Ls
 Ls 


Solving s-Domain Circuit Equations
G + 1 Ls
− 1 Ls
• Circuit Determinant: ∆( s ) =
− 1 Ls Cs + 1 Ls
= (G + 1 Ls )(Cs + 1 Ls ) − (1 Ls ) 2
GLCs 2 + Cs + G
=
Ls
Depends on circuit element parameters: L, C, G=1/R,
not on driving force and initial conditions
• Solve for node A using Cramer’s rule:
− 1 Ls
I S ( s ) + i L ( 0) s

∆ A ( s) iL (0) s + CvC (0) Cs + 1 Ls
=
VA ( s ) =
∆(s)
∆(s)

( LCs 2 + 1) I S ( s) − LCsiL (0) + CvC (0)
=
+
2
GLCs + Cs + G
GLCs 2 + Cs + G
Zero State
when initial condition
sources are turned off

Zero input
when input sources
are turned off


Solving s-Domain Circuit Eqns. (Cont’d)
• Solve for node B using Cramer’s rule:

G + 1 Ls

I S ( s ) − iL (0) s

− 1 Ls iL (0) s + CvC (0)
∆ B (s)
=
∆( s)
∆( s)
I S ( s)
GLiL (0) + (GLs + 1)CvC (0)
=

+
2
GLCs + Cs + G
GLCs 2 + Cs + G

VB ( s ) =

Zero State

Zero input


Network Functions
Network function =

Zero - state Response Transform
Input Signal Transform

• Driving-point function relates the voltage and
current at a given pair of terminals called a port
V (s)
1
=
Z ( s) =
I ( s) Y ( s)
• Transfer function relates an input and response at
different ports in the circuit
V (s)
TV ( s ) = Voltage Transfer Function = 2
V1 ( s )

I 2 (s)
TI ( s ) = Current Transfer Function =
V1
I1 ( s )
I ( s)
In
TY ( s) = Transfer Admittance = 2
V1 ( s)
V (s)
V1
TZ ( s) = Transfer Impedance = 2
I1 ( s )
_
+
_
+

In

I (s )

+

Circuit
in the
zero-state

V (s )
−

Circuit Output
in the
V1 or I1
V or I
zero-state 2 2
Input

I2

+
V2
−

TV (s )
Out

I1 TI (s )
In

Out

I2
TY (s )

+
V2
−

I1 TZ (s )
Out

In

Out


Calculating Network Functions
Z1
V1 ( s )

+
Z2 V2 ( s)
−

I 2 (s)
I1 ( s )

Y1

Y2

• Driving-point impedance
Z EQ ( s ) = Z1 ( s) + Z 2 ( s )
• Voltage transfer function:


Z 2 (s)
V2 ( s ) = 
V1 ( s )
 Z1 ( s ) + Z 2 ( s ) 

V (s)
Z 2 ( s)
TV ( s) = 2
=
V1 ( s ) Z1 ( s ) + Z 2 ( s )

• Driving-point admittance
YEQ ( s ) = Y1 ( s ) + Y2 ( s)
• Voltage transfer function:
 Y2 ( s )

I 2 (s) = 
Y1 ( s)
 Y1 ( s) + Y2 ( s ) 
I (s)
Y2 ( s )
TI ( s ) = 2
=
I1 ( s ) Y1 ( s ) + Y2 ( s)

_
+


Impulse Response and Step Response
• Input-output relationship in s-domain
Input
Y ( s) = T ( s) X ( s)
X (s )
• When input signal is an impulse x(t ) = δ (t )

Y ( s) = T ( s) ×1 = T ( s)

T(s)
Circuit

Output

Y (s )

– Impulse response equals network function
– H(s) = impulse response transform
– h(t) = impulse response waveform

• When input signal is a step x(t ) = u (t )
– G(s) = step response transform
– g(t) = step response waveform

T (s) H (s)
=
G (s) =
s
s
g ( s) = ∫ h(τ )dτ ,
t

0

(=) means equal almost everywhere,

dg (t )

excludes those points at which g(t)
h(t )(=)
dt
has a discontinuity